Electronic Journal of Differential Equations, Vol. 2006(2006), No. 117, pp. 1–11.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
EXISTENCE AND UNIQUENESS OF PERIODIC SOLUTIONS
FOR FIRST-ORDER NEUTRAL FUNCTIONAL DIFFERENTIAL
EQUATIONS WITH TWO DEVIATING ARGUMENTS
JINSONG XIAO, BINGWEN LIU
Abstract. In this paper, we use the coincidence degree theory to establish
the existence and uniqueness of T -periodic solutions for the first-order neutral
functional differential equation, with two deviating arguments,
(x(t) + Bx(t − δ))0 = g1 (t, x(t − τ1 (t))) + g2 (t, x(t − τ2 (t))) + p(t).
1. Introduction
Consider the first-order neutral functional differential equation (NFDE), with
two deviating arguments,
(x(t) + Bx(t − δ))0 = g1 (t, x(t − τ1 (t))) + g2 (t, x(t − τ2 (t))) + p(t),
(1.1)
where τ1 , τ2 , p : R → R and g1 , g2 : R × R → R are continuous functions, B and
δ are constants, τ1 , τ2 and p are T -periodic, g1 and g2 are T -periodic in the first
argument, |B| =
6 1 and T > 0.
The above equation has been used for the study of distributed networks containing lossless transmission lines [6, 7]. Hence, in recent years, the problem of
the existence of periodic solutions for (1.1) has been extensively studied. For more
details, we refer the reader to [1, 2, 4, 5, 6, 7, 9, 12] and the references cited therein.
However, to the best of our knowledge, there exist no results for the existence and
uniqueness of periodic solutions of (1.1).
The main purpose of this paper is to establish sufficient conditions for the existence and uniqueness of T -periodic solutions of (1.1). The results of this paper
are new and they complement previously known results. An illustrative example is
given in Section 4.
For ease of exposition, throughout this paper we will adopt the following notation:
Z T
1/k
|x|k =
|x(t)|k dt
, |x|∞ = max |x(t)|.
t∈[0,T ]
0
2000 Mathematics Subject Classification. 34C25, 34D40.
Key words and phrases. First order; neutral; functional differential equations;
deviating argument; periodic solutions; coincidence degree.
c
2006
Texas State University - San Marcos.
Submitted March 16, 2006. Published September 26, 2006.
Supported by grants 10371034 from the NNSF of China and 05JJ40009 from the
Hunan Provincial Natural Science Foundation of China.
1
2
J. XIAO,B. LIU
EJDE-2006/117
Let X = {x|x ∈ C(R, R), x(t + T ) = x(t), for all t ∈ R} be a Banach space with
the norm kxkX = |x|∞ . Define the two linear operators
A : X → X,
(Ax)(t) = x(t) + Bx(t − δ);
L : D(L) ⊂ X → X,
Lx = (Ax)0 ,
(1.2)
where D(L) = {x|x ∈ X, x0 ∈ C(R, R)}.
We also define the nonlinear operator N : X → X by
N x = g1 (t, x(t − τ1 (t))) + g2 (t, x(t − τ2 (t))) + p(t).
By Hale’s terminology [4], a solution u(t) of (1.1) is that u ∈ C(R, R) such that
Au ∈ C 1 (R, R) and (1.1) is satisfied on R. In general, u 6∈ C 1 (R, R). But from [9,
Lemma 1], in view of |B| 6= 1, it is easy to see that (Ax)0 = Ax0 . So a T -periodic
solution u(t) of (1.1) must be such that u ∈ C 1 (R, R). Meanwhile, according to [9,
RT
Lemma 1], we can easily get that ker L = R, and Im L = {x ∈ X : 0 x(s)ds = 0}.
Therefore, the operator L is a Fredholm operator with index zero. Define the
continuous projectors P : X → ker L and Q : X → X/ImL by setting
Z
1 T
x(s)ds,
P x(t) =
T 0
Z
1 T
Qx(t) =
x(s)ds.
T 0
Hence, Im P = ker L and ker Q = Im L. Set LP = L|D(L)∩KerP , then LP has
continuous inverse L−1
P defined by
Z t
Z T
−1 1
L−1
y(t)
=
A
sy(s)ds
+
y(s)ds
.
(1.3)
P
T 0
0
Therefore, it is easy to see from (1) and (1.3) that N is L-compact on Ω, where Ω
is an open bounded set in X.
2. Preliminary Results
In view of (1.2) and (1), the operator equation
Lx = λN x
is equivalent to the equation
x0 (t) + Bx0 (t − δ) = λ[g1 (t, x(t − τ1 (t))) + g2 (t, x(t − τ2 (t))) + p(t)],
where λ ∈ (0, 1).
For convenience of use, we introduce the Continuation Theorem [2] as follows.
Lemma 2.1. Let X be a Banach space. Suppose that L : D(L) ⊂ X → X is a
Fredholm operator with index zero and N : Ω → X is L-compact on Ω, where Ω is
an open bounded subset of X. Moreover, assume that all the following conditions
are satisfied:
(1) Lx 6= λN x, for all x ∈ ∂Ω ∩ D(L), λ ∈ (0, 1);
(2) N x 6∈ ImL, for all x ∈ ∂Ω ∩ ker L;
(3) For the Brower degree, deg{QN, Ω ∩ ker L, 0} =
6 0.
Then the equation Lx = N x has at least one solution on Ω ∩ D(L).
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EXISTENCE AND UNIQUENESS OF PERIODIC SOLUTIONS
3
By using a similar argument of the proof of [8, Lemma 2.5], from [3, Theorem
225], we can obtain the following Lemma.
Lemma 2.2. Let x(t) ∈ X ∩ C 1 (R, R) . Suppose that there exists a constant D ≥ 0
such that
|x(τ0 )| ≤ D, τ0 ∈ [0, T ].
Then
√
T
|x|2 ≤ |x0 |2 + T D.
(2.1)
π
Lemma 2.3 ([10]). Let µ ∈ [0, T ] be a constant, δ ∈ C(R, R) be periodic with
period T , and supt∈[0,T ] |δ(t)| ≤ µ. Then for any h ∈ C 1 (R, R) which is periodic
with period T , we have
Z T
Z T
2
2
|h(s) − h(s − δ(s))| ds ≤ 2µ
|h0 (s)|2 ds.
(2.2)
0
0
For the next lemma we need the following conditions
(H) For i = 1, 2, there exist a constants µi and an integers Ki such that
µi = sup |τi (t) − Ki T | ≤ T.
t∈[0,T ]
(A0) One of the following conditions holds:
(1) (gi (t, u1 ) − gi (t, u2 ))(u1 − u2 ) > 0, for i = 1, 2, ui ∈ R, for all t ∈ R and
u1 6= u2 ;
(2) (gi (t, u1 ) − gi (t, u2 ))(u1 − u2 ) < 0, for i = 1, 2, ui ∈ R, for all t ∈ R and
u1 6= u2 ;
(A0’) One of the following conditions holds:
√
√
(1) there exists constants b1 and b2 such that b1 ( 2µ1 + Tπ )+b2 ( 2µ2 + Tπ ) <
1 − |B|, and
|gi (t, u1 ) − gi (t, u2 )| ≤ bi |u1 − u2 |, for i = 1, 2, ui ∈ R, ∀t ∈ R,
√
√
(2) There exists constants b1 and b2 such that b1 ( 2µ1 + Tπ ) + b2 ( 2µ2 +
T
π ) < |B| − 1, and
|gi (t, u1 ) − gi (t, u2 )| ≤ bi |u1 − u2 |,
for i = 1, 2, ui ∈ R, ∀t ∈ R.
Lemma 2.4. Under assumptions (A0) and (A0’), Equation (1.1) has at most one
T -periodic solution.
Proof. Suppose that x1 (t) and x2 (t) are two T -periodic solutions of (1.1). Then
(x1 (t) + Bx1 (t − δ))0 − g1 (t, x1 (t − τ1 (t))) − g2 (t, x1 (t − τ2 (t))) = p(t)
and
(x2 (t) + Bx2 (t − δ))0 − g1 (t, x2 (t − τ1 (t))) − g2 (t, x2 (t − τ2 (t))) = p(t).
This implies
[(x1 (t) − x2 (t)) + B(x1 (t − δ) − x2 (t − δ))]0 − (g1 (t, x1 (t − τ1 (t)))
− g1 (t, x2 (t − τ1 (t)))) − (g2 (t, x1 (t − τ2 (t))) − g2 (t, x2 (t − τ2 (t)))) = 0.
(2.3)
Set Z(t) = x1 (t) − x2 (t). Then, from (2.3), we obtain
Z 0 (t) + BZ 0 (t − δ) − (g1 (t, x1 (t − τ1 (t)))
− g1 (t, x2 (t − τ1 (t)))) − (g2 (t, x1 (t − τ2 (t))) − g2 (t, x2 (t − τ2 (t)))) = 0.
(2.4)
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J. XIAO,B. LIU
EJDE-2006/117
Thus, integrating (2.4) from 0 to T , we have
Z T
[(g1 (t, x1 (t − τ1 (t))) − g1 (t, x2 (t − τ1 (t))))
0
+ (g2 (t, x1 (t − τ2 (t))) − g2 (t, x2 (t − τ2 (t))))]dt = 0.
Therefore, in view of integral mean value theorem, it follows that there exists a
constant γ ∈ [0, T ] such that
(g1 (γ, x1 (γ − τ1 (γ))) − g1 (γ, x2 (γ − τ1 (γ))))
+ (g2 (γ, x1 (γ − τ2 (γ))) − g2 (γ, x2 (γ − τ2 (γ)))) = 0.
(2.5)
From (A0), (2.5) implies
(x1 (γ − τ1 (γ)) − x2 (γ − τ1 (γ)))(x1 (γ − τ2 (γ)) − x2 (γ − τ2 (γ))) ≤ 0.
Since Z(t) = x1 (t) − x2 (t) is a continuous function on R, it follows that there exists
a constant ξ ∈ R such that
Z(ξ) = 0.
(2.6)
Let ξ = nT + γ
e, where γ
e ∈ [0, T ] and n is an integer. Then, (2.6) implies that there
exists a constant γ
e ∈ [0, T ] such that
Z(e
γ ) = Z(ξ) = 0.
(2.7)
Then, from Lemma 2.2, using Schwarz inequality and the inequality
Z t
Z T
|Z(t)| = |Z(e
γ) +
Z 0 (s)ds| ≤
|Z 0 (s)|ds, for all t ∈ [0, T ],
γ
e
we obtain
|Z|∞ ≤
0
√
T |Z 0 |2 ,
and |Z|2 ≤
T 0
|Z |2 .
π
(2.8)
Now, we consider two cases.
Case (i). If (A0’)(1) holds, multiplying both sides of (2.4) by Z 0 (t) and then
integrating them from 0 to T , using (H), (2.2), (2.8) and Schwarz inequality, we
have
|Z 0 |22
Z
=
T
|Z 0 (t)|2 dt
0
T
Z
0
= −B
Z
0
Z
Z (t)Z (t − δ)dt +
0
T
T
(g1 (t, x1 (t − τ1 (t))) − g1 (t, x2 (t − τ1 (t))))Z 0 (t)dt
0
(g2 (t, x1 (t − τ2 (t))) − g2 (t, x2 (t − τ2 (t))))Z 0 (t)dt
+
0
≤ |B||Z 0 |22 + b1
Z
T
|x1 (t − τ1 (t)) − x2 (t − τ1 (t))||Z 0 (t)|dt
0
Z
T
|x1 (t − τ2 (t)) − x2 (t − τ2 (t))||Z 0 (t)|dt
+ b2
0
≤
|B||Z 0 |22
Z
+ b1
T
|Z(t − τ1 (t)) − Z(t)||Z 0 (t)|dt + b1
0
Z
+ b2
0
T
|Z(t − τ2 (t)) − Z(t)||Z 0 (t)|dt + b2
Z
T
|Z(t)||Z 0 (t)|dt
0
Z
0
T
|Z(t)||Z 0 (t)|dt
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EXISTENCE AND UNIQUENESS OF PERIODIC SOLUTIONS
T
Z
≤ |B||Z 0 |22 + b1 (
5
1
|Z(t − τ1 (t)) − Z(t)|2 dt) 2 |Z 0 |2 + b1 |Z|2 |Z 0 |2
0
T
Z
+ b2 (
1
|Z(t − τ2 (t)) − Z(t)|2 dt) 2 |Z 0 |2 + b2 |Z|2 |Z 0 |2
0
T
Z
= |B||Z 0 |22 + b1 (
1
|Z(t − (τ1 (t) − K1 T )) − Z(t)|2 dt) 2 |Z 0 |2 + b1 |Z|2 |Z 0 |2
0
T
Z
+ b2 (
1
|Z(t − (τ2 (t) − K2 T )) − Z(t)|2 dt) 2 |Z 0 |2 + b2 |Z|2 |Z 0 |2
0
√
√
T
T
≤ [|B| + b1 ( 2µ1 + ) + b2 ( 2µ2 + )]|Z 0 |22 .
π
π
From (2.8) and (A0’)(1), the above inequalit implies
Z(t) ≡ Z 0 (t) ≡ 0,
for all t ∈ R.
Hence, x1 (t) ≡ x2 (t), for all t ∈ R. Therefore, (1.1) has at most one T -periodic
solution.
Case (ii). If (A0’)(2) holds, multiplying both sides of (2.4) by Z 0 (t − δ) and then
integrating them from 0 to T , using (H), (2.2), (2.8) and Schwarz inequality, we
have
|B||Z 0 |22
Z T
=|
B|Z 0 (t − δ)|2 dt|
0
T
Z
Z 0 (t)Z 0 (t − δ)dt
=|−
Z
0
T
(g1 (t, x1 (t − τ1 (t))) − g2 (t, x2 (t − τ1 (t))))Z 0 (t − δ)dt
+
0
Z
T
(g2 (t, x1 (t − τ2 (t))) − g2 (t, x2 (t − τ2 (t))))Z 0 (t − δ)dt|
+
0
≤ |Z 0 |22 + b1
T
Z
|x1 (t − τ1 (t)) − x2 (t − τ1 (t))||Z 0 (t − δ)|dt
0
T
Z
|x1 (t − τ2 (t)) − x2 (t − τ2 (t))||Z 0 (t − δ)|dt
+ b2
0
≤
|Z 0 |22
T
Z
Z
0
|Z(t − τ1 (t)) − Z(t)||Z (t − δ)|dt + b1
+ b1
0
|Z(t)||Z 0 (t − δ)|dt
0
T
Z
|Z(t − τ2 (t)) − Z(t)||Z 0 (t − δ)|dt + b2
+ b2
T
0
T
Z
|Z(t)||Z 0 (t − δ)|dt
0
Z
≤ |Z 0 |22 + b1 (
T
1
|Z(t − τ1 (t)) − Z(t)|2 dt) 2 |Z 0 |2 + b1 |Z|2 |Z 0 |2
0
Z
+ b2 (
T
1
|Z(t − τ2 (t)) − Z(t)|2 dt) 2 |Z 0 |2 + b2 |Z|2 |Z 0 |2
0
=
|Z 0 |22
Z
+ b1 (
0
T
1
|Z(t − (τ1 (t) − K1 T )) − Z(t)|2 dt) 2 |Z 0 |2 + b1 |Z|2 |Z 0 |2
6
J. XIAO,B. LIU
Z
+ b2 (
EJDE-2006/117
T
1
|Z(t − (τ2 (t) − K2 T )) − Z(t)|2 dt) 2 |Z 0 |2 + b2 |Z|2 |Z 0 |2
0
√
√
T
T
≤ [1 + b1 ( 2µ1 + ) + b2 ( 2µ2 + )]|Z 0 |22
π
π
Then using the methods similar to those used in Case (i), from the above inequality,
(2.8), and (A0’)(2), we can conclude that (1.1) has at most one T -periodic solution.
The proof of Lemma 2.4 is now complete.
For the next lemma we use the following assumptions:
(A1) x(g1 (t, x) + g2 (t, x) + p(t)) > 0, for all t ∈ R, |x| ≥ d;
(A2) x(g1 (t, x) + g2 (t, x) + p(t)) < 0, for all t ∈ R, |x| ≥ d.
Lemma 2.5. Assume (A0) and that there exists a positive constant d such that
one of the two conditions (A1) or (A2) holds. If x(t) is a T -periodic solution of
(2), then
√
(2.9)
|x|∞ ≤ d + T |x0 |2 .
Proof. Let x(t) be a T -periodic solution of (2). Then, integrating (2) from 0 to T ,
we have
Z T
[g1 (t, x(t − τ1 (t))) + g2 (t, x(t − τ2 (t))) + p(t)]dt = 0.
0
This implies that there exists a constant t1 ∈ R such that
g1 (t1 , x(t1 − τ1 (t1 ))) + g2 (t1 , x(t1 − τ2 (t1 ))) + p(t1 ) = 0.
(2.10)
We show next the following Claim: If x(t) is a T -periodic solution of (2), then there
exists a constant t2 ∈ R such that
|x(t2 )| ≤ d.
(2.11)
Assume, by way of contradiction, that (2.11) does not hold. Then
|x(t)| > d,
for all t ∈ R,
which, together with (A1), (A2) and (2.10), implies that one of the following relations holds:
x(t1 − τ1 (t1 )) > x(t1 − τ2 (t1 )) > d;
(2.12)
x(t1 − τ2 (t1 )) > x(t1 − τ1 (t1 )) > d;
(2.13)
x(t1 − τ1 (t1 )) < x(t1 − τ2 (t1 )) < −d;
(2.14)
x(t1 − τ2 (t1 )) < x(t1 − τ1 (t1 )) < −d.
(2.15)
If (2.12) holds, in view of (A0)(1), (A0)(2), (A1) and (A2), we shall consider four
cases as follows.
Case (i). If (A1) and (A0)(1) hold, according to (2.12), we obtain
0 < g1 (t1 , x(t1 − τ2 (t1 ))) + g2 (t1 , x(t1 − τ2 (t1 ))) + p(t1 )
< g1 (t1 , x(t1 − τ1 (t1 ))) + g2 (t1 , x(t1 − τ2 (t1 ))) + p(t1 ),
which contradicts (2.10). This contradiction implies that (2.11) holds.
Case (ii). If (A1) and (A0)(2) hold, according to (2.12), we obtain
0 < g1 (t1 , x(t1 − τ1 (t1 ))) + g2 (t1 , x(t1 − τ1 (t1 ))) + p(t1 )
< g1 (t1 , x(t1 − τ1 (t1 ))) + g2 (t1 , x(t1 − τ2 (t1 ))) + p(t1 ),
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EXISTENCE AND UNIQUENESS OF PERIODIC SOLUTIONS
7
which contradicts (2.10). This contradiction implies that (2.11) holds.
Case (iii). If (A2) and (A0)(1) hold, according to (2.12), we obtain
g1 (t1 , x(t1 − τ1 (t1 ))) + g2 (t1 , x(t1 − τ2 (t1 ))) + p(t1 )
< g1 (t1 , x(t1 − τ1 (t1 ))) + g2 (t1 , x(t1 − τ1 (t1 ))) + p(t1 ) < 0,
which contradicts (2.10). This contradiction implies that (2.11) holds.
Case (iv). If (A2) and (A0)(2) hold, according to (2.12), we obtain
g1 (t1 , x(t1 − τ1 (t1 ))) + g2 (t1 , x(t1 − τ2 (t1 ))) + p(t1 )
< g1 (t1 , x(t1 − τ2 (t1 ))) + g2 (t1 , x(t1 − τ2 (t1 ))) + p(t1 ) < 0,
which contradicts (2.10). This contradiction implies that (2.11) holds.
If (2.13) (or (2.14), or (2.15)) holds, using the methods similar to those used in
Case (i) - Case (iv), we can show that (2.11) holds. This completes the proof of
the Claim.
Let t2 = mT + t0 , where t0 ∈ [0, T ] and m is an integer. Then, using Schwarz
inequality and the inequality
Z
t
|x(t)| = |x(t0 ) +
x0 (s)ds| ≤ d +
t0
Z
T
|x0 (s)|ds,
for all t ∈ [0, T ],
0
we obtain
|x|∞ = max |x(t)| ≤ d +
√
t∈[0,T ]
This completes the proof.
T |x0 |2 .
3. Main Results
Theorem 3.1. Assume that (H), (A0), (A0’) and either (A1) or (A2). Then (1.1)
has a unique T -periodic solution.
Proof. From Lemma 2.4, together with (H), (A0) and (A0’), it i s easy to see that
(1.1) has at most one T -periodic solution. Thus, to prove Theorem 3.1, it suffices
to show that (1.1) has at least one T -periodic solution. To do this, we shall apply
Lemma 2.1. Firstly, we will claim that the set of all possible T -periodic solutions
of (2) is bounded.
Let x(t) be a T -periodic solution of equation (2). In view of (A0’)(1) and
(A0’)(2), we shall consider two cases as follows.
Case (i). If (A0’)(1) holds, multiplying both sides of (2) by x0 (t) and then integrating them from 0 to T , from (2.1), (2.2), (2.11), (H), (A0’)(1) and the Schwarz
8
J. XIAO,B. LIU
EJDE-2006/117
inequality, we have
|x0 |22
Z
=
T
|x0 (t)|2 dt
0
T
Z
0
Z
0
Bx (t − δ)x (t)dt + λ
=−
T
g1 (t, x(t − τ1 (t)))x0 (t)dt
0
0
Z
T
+λ
g2 (t, x(t − τ2 (t)))x0 (t)dt + λ
0
Z
T
p(t)x0 (t)dt
0
(g1 (t, x(t − τ1 (t))) − g1 (t, x(t)) + g1 (t, x(t))
0
T
Z
0
T
Z
≤ |B||x0 |22 + |p|2 |x0 |2 + λ
− g1 (t, 0))x (t)dt + λ
(g2 (t, x(t − τ2 (t))) − g2 (t, x(t)) + g2 (t, x(t))
0
T
Z
0
Z
0
− g2 (t, 0))x (t)dt + λ
g1 (t, 0)x (t)dt + λ
0
T
g2 (t, 0)x0 (t)dt
0
T
Z
|B||x0 |22 + |p|2 |x0 |2 + b1 (
1
|x(t − τ1 (t)) − x(t)|2 dt) 2 |x0 |2 + b1 |x|2 |x0 |2
0
Z
+ b2 (
T
1
|x(t − τ2 (t)) − x(t)|2 dt) 2 |x0 |2 + b2 |x|2 |x0 |2
0
√
+ ( max |g1 (t, 0)| + max |g2 (t, 0)|) T |x0 |2
t∈[0,T ]
t∈[0,T ]
Z
= |B||x0 |22 + |p|2 |x0 |2 + b1 (
T
1
|x(t − (τ1 (t) − K1 T )) − x(t)|2 dt) 2 |x0 |2 + b1 |x|2 |x0 |2
0
Z
+ b2 (
T
1
|x(t − (τ2 (t) − K2 T )) − x(t)|2 dt) 2 |x0 |2 + b2 |x|2 |x0 |2
0
√
+ ( max |g1 (t, 0)| + max |g2 (t, 0)|) T |x0 |2
t∈[0,T ]
t∈[0,T ]
√
√
T
T
≤ [|B| + b1 ( 2µ1 + ) + b2 ( 2µ2 + )]|x0 |22 + |p|2 |x0 |2
π
π
√
+ (b1 d + b2 d + max |g1 (t, 0)| + max |g2 (t, 0)|) T |x0 |2 .
t∈[0,T ]
t∈[0,T ]
(3.1)
Now, let
√
|p|2 + (b1 d + b2 d + maxt∈[0,T ] |g1 (t, 0)| + maxt∈[0,T ] |g2 (t, 0)|) T
√
√
D1 =
.
1 − |B| − (b1 ( 2µ1 + Tπ ) + b2 ( 2µ2 + Tπ ))
In view of (2.9) and (3.1), we obtain
|x0 |2 ≤ D1 , |x|∞ ≤ d +
√
T D1 .
(3.2)
Case (ii). If (A0’)(2) holds, multiplying both sides of (2) by x0 (t − δ) and then
integrating them from 0 to T , from (2.1), (2.2), (2.9), (2.11), (A0’)(2) and the
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EXISTENCE AND UNIQUENESS OF PERIODIC SOLUTIONS
9
inequality of Schwarz, we have
Z T
B|x0 (t − δ)|2 dt|
|B||x0 |22 = |
0
T
Z
0
T
Z
0
g1 (t, x(t − τ1 (t)))x0 (t − δ)dt
x (t − δ)x (t)dt + λ
=|−
0
0
Z
T
+λ
g2 (t, x(t − τ2 (t)))x0 (t − δ)dt + λ
Z
0
T
p(t)x0 (t − δ)dt|
0
≤ |x0 |22 + |p|2 |x0 |2 + |λ
T
Z
(g1 (t, x(t − τ1 (t))) − g1 (t, x(t)) + g1 (t, x(t))
0
− g1 (t, 0))x0 (t − δ)dt + λ
Z
T
(g2 (t, x(t − τ2 (t))) − g2 (t, x(t)) + g2 (t, x(t))
0
Z
0
− g2 (t, 0))x (t − δ)dt + λ
T
Z
0
0
≤
|x0 |22
Z
+ |p|2 |x |2 + b1 (
0
T
g1 (t, 0)x (t)dt + λ
g2 (t, 0)x0 (t − δ)dt|
0
T
1
|x(t − τ1 (t)) − x(t)|2 dt) 2 |x0 |2 + b1 |x|2 |x0 |2
0
Z
+ b2 (
T
1
|x(t − τ2 (t)) − x(t)|2 dt) 2 |x0 |2 + b2 |x|2 |x0 |2
0
√
+ ( max |g1 (t, 0)| + max |g2 (t, 0)|) T |x0 |2
t∈[0,T ]
t∈[0,T ]
√
√
T
T
≤ [1 + b1 ( 2µ1 + ) + b2 ( 2µ2 + )]|x0 |22 + |p|2 |x0 |2
π
π
√
+ (b1 d + b2 d + max |g1 (t, 0)| + max |g2 (t, 0)|) T |x0 |2 .
t∈[0,T ]
t∈[0,T ]
(3.3)
Now, let
√
|p|2 + (b1 d + b2 d + maxt∈[0,T ] |g1 (t, 0)| + maxt∈[0,T ] |g2 (t, 0)|) T
√
√
D1 =
.
|B| − 1 − (b1 ( 2µ1 + Tπ ) + b2 ( 2µ2 + Tπ ))
In view of (2.9) and (3.3), we obtain
|x0 |2 ≤ D1 , |x|∞ ≤ d +
√
T D1 .
(3.4)
If x ∈ Ω1 = {x ∈ ker L ∩ X : N x ∈ ImL}, then there exists a constant M1 such
that
Z T
x(t) ≡ M1 and
[g1 (t, M1 ) + g2 (t, M1 ) + p(t)]dt = 0.
(3.5)
0
Thus,
|x(t)| ≡ |M1 | < d,
√
Let M = (D1 + D1 ) T + d + 1. Set
for all x(t) ∈ Ω1 .
(3.6)
Ω = {x|x ∈ X, |x|∞ < M }.
It is easy to see from (1) and (1.3)
√ that N is√L-compact on Ω. We have from (3.5),
(3.6) and the fact M > max{D1 T + d, D1 T + d, d} that the conditions (1) and
(2) in Lemma 2.1 hold.
10
J. XIAO,B. LIU
EJDE-2006/117
Furthermore, define the continuous functions
Z
1 T
H1 (x, µ) = (1 − µ)x + µ ·
[g1 (t, x) + g2 (t, x) + p(t)]dt; µ ∈ [0 1],
T 0
Z
1 T
H2 (x, µ) = −(1 − µ)x + µ ·
[g1 (t, x) + g2 (t, x) + p(t)]dt; µ ∈ [0 1].
T 0
If (A1) holds, then
xH1 (x, µ) 6= 0 for allx ∈ ∂Ω ∩ ker L.
Hence, using the homotopy invariance theorem, we have
Z
1 T
deg{QN, Ω ∩ ker L, 0} = deg{
[g1 (t, x) + g2 (t, x) + p(t)]dt, Ω ∩ ker L, 0}
T 0
= deg{x, Ω ∩ ker L, 0} =
6 0.
If (A2) holds, then xH2 (x, µ) 6= 0 for all x ∈ ∂Ω∩ker L. Hence, using the homotopy
invariance theorem, we obtain
Z
1 T
[g1 (t, x) + g2 (t, x) + p(t)]dt, Ω ∩ ker L, 0}
deg{QN, Ω ∩ ker L, 0} = deg{
T 0
= deg{−x, Ω ∩ ker L, 0} =
6 0.
In view of all the discussions above and Lemma 2.1, Theorem 3.1 is proved.
4. Concluding remarks
Example 4.1. The first-order neutral functional differential
√
√
3
1
1
2
2
2
0
sin t) + [1 − x(t −
cos2 t)] + ecos t (4.1)
(x(t) + x(t − δ)) = − x(t −
8
8
64
32
64
has a unique 2π-periodic solution.
1
From (4.1), we have B = 18 , g1 (x) = − 38 x, g2 (x) = 32
[1 − x] and p(t) = ecos t .
√
√
√
√
Then, µ1 = supt∈[0,T ] | 642 sin2 t| = 642 < 2π, µ2 = supt∈[0,T ] | 642 cos2 t| = 642 < 2π,
1
b1 = 38 , b2 = 32
. It is straight forward to check that all the conditions needed in
Theorem 3.1 are satisfied. Therefore, (4.1) has a unique 2π-periodic solution.
Remark 4.2. Equation (4.1) is a very simple version of first order NFDE. Since
B 6= 0, all the results in the references and their references can not be applicable to
(4.1) to obtain the existence and uniqueness of 2π-periodic solutions. This implies
that the results of this paper are essentially new.
Remark 4.3. By using the methods similarly to those used for (1.1), we can deal
with the NFDE with multiple deviating arguments, for example
(x(t) + Bx(t − δ))0 =
n
X
gi (t, x(t − τi (t))) + p(t),
(4.2)
i=1
where τi (i = 1, 2, . . . , n), p : R → R and gi (i = 1, 2, . . . , n) : R × R → R are
continuous functions, τi (i = 1, 2, . . . , n) and p are T -periodic, gi , i = 1, 2, . . . , n, are
T -periodic in the first argument, and T > 0. One may also establish the results
similarly to those in Theorem 3.1 under some minor additional assumptions on
gi (t, x)(i = 1, 2, . . . , n).
EJDE-2006/117
EXISTENCE AND UNIQUENESS OF PERIODIC SOLUTIONS
11
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Jinsong Xiao
Department of Mathematics and Computer Science, Hunan City University, Yiyang,
Hunan 413000, China
E-mail address: bingxiao209@yahoo.com.cn
Bingwen Liu
College of Mathematics and Information Science, Jiaxing University, Jiaxing, Zhejiang
314001, China
E-mail address: liubw007@yahoo.com.cn